show that the 4 points (1,0),(9,-6),(2,-7) and (8,1) are coincyclic points

The standard equation of the of a circle x2+y2+dx+ey+f=0

Now put points  (1,0) ,(2,-7) ,(9,-6) in equations  of circle, we get three equations respectively.

1+d+f=0 , 

2d-7e+f=-53 ,

9d-6e+f=-117

Now after  solving these equations we get d= -10, e= 6, f=9 and now for the proof of the concyclicity we have to form the standard equation and then we have to check whether the fourth point satisfy the standard equation or not if it will satisfy the equation then these points are concyclic otherwise not.
Thus eqn. of circle is x2+y2-10x+6y+9=0
Put (8,1) in this equation =82+12-10*8+6*1+9
= 0

And we find it satisfy the eqn of circle. Thus all points are coincyclic.

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we know eqn- of circle given as x2+y2+2gx+2fy+c. one by one put these points in the eqn- given satisfy the last point in this eqn-. I am sure you will get answer.

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